Question Number 217255 by MrGaster last updated on 07/Mar/25
$$\int_{−\infty} ^{+\infty} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx}=\sqrt{\mathrm{2}\pi},\int_{−\infty} ^{+\infty} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{x}} {dx}. \\ $$
Answered by vnm last updated on 07/Mar/25
$$={e}^{\frac{\mathrm{1}}{\mathrm{2}}} \underset{−\infty} {\overset{+\infty} {\int}}{e}^{−\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}} \mathrm{d}\left({x}−\mathrm{1}\right)=\sqrt{\mathrm{2}\pi{e}}. \\ $$